Let $f: A \to B$ a faithful flat map between rings. Let $a \in A$.
I want to know how to prove that
$$aA = aB \cap A$$ holds.
Ideas:
Consider the exact sequence
(1) $$0 \to aB \cap A\to A \to A/(aB \cap A)\to 0$$
Tensoring by $B$ gives exact
(2) $$0 \to (aB \cap A) \otimes_A B \to B \to A/(aB \cap A) \otimes_A B \to 0$$
Does here $(aB \cap A) \otimes_A B= aB$ hold?
My idea is to show that $aA = \operatorname{Ker}(A\to A \to A/(aB \cap A)) =aB \cap A$
by going back from (2) to (1). Since $aB = aA \otimes B$ then by faithful flatness that suffice.
But I need $(aB \cap A) \otimes_A B= aB$.
Or is this generally wrong and I should try another idea?